# Thread: What are the Zero's of this function?

1. ## What are the Zero's of this function?

the problem is {x}^{4} - 16

I get;
{x}^{4} - {2}^{4} = {(x+2)}^{2} + {(x-2)}^{2}

for zeros of 2 and -2 each of multiplicity of 2.

However the book states zeros of 2 and -2 each of multiplicity 1. Could someone be so kind as to explain to me what I have done wrong?

2. Originally Posted by bkbowser
the problem is {x}^{4} - 16

I get;
{x}^{4} - {2}^{4} = {(x+2)}^{2} + {(x-2)}^{2}

for zeros of 2 and -2 each of multiplicity of 2.

However the book states zeros of 2 and -2 each of multiplicity 1. Could someone be so kind as to explain to me what I have done wrong?
You can applied the difference of square incorrectly.

$\displaystyle x^4-16=(x^2)^2-4^2=(x^2-4)(x^2+4)$

Now use the difference of squares again!

3. Originally Posted by TheEmptySet
You can applied the difference of square incorrectly.

$\displaystyle x^4-16=(x^2)^2-4^2=(x^2-4)(x^2+4)$

Now use the difference of squares again!
$\displaystyle (x^2+4)(x-2)(x+2)$ Is this it? It does give zeros of 2 and -2 each of multiplicity 1. And since the question just asked about real zeros the other 2 wouldn't count?

4. Originally Posted by bkbowser
$\displaystyle (x^2+4)(x-2)(x+2)$ Is this it? It does give zeros of 2 and -2 each of multiplicity 1. And since the question just asked about real zeros the other 2 wouldn't count?
yes

5. Another way to solve $\displaystyle x^4- 16= 0$ is to write it as $\displaystyle x^4= 16$. Taking the square root of each side we have $\displaystyle x^2= 4$ and $\displaystyle x^2= -4$. From $\displaystyle x^2= 4$, taking the square root again, x= 2 and x= -2. There are no real number x such that $\displaystyle x^2= -4$.